A set $S = \{\mathbf{x_1 \space \dots \space \mathbf{x_k} }\}$ with nonzero elements $\mathbf{x_i} \neq \mathbf{0}$ is called orthogonal if
$$ \forall i, j \in \{0 \space \dots \space k\}, \space i\neq j, \hspace{10pt} \langle \mathbf{x_i} \space , \mathbf{x_j} \rangle = \mathbf{x_i}^\text{T} \mathbf{x_j}^* = 0 $$
If $S$ is orthogonal:
$S$ is also linearly independent. The converse is not generally true.
Coefficients to the representation of any vector within its span can be found succinctly.
$$ \sum _{i=1} ^k \alpha_i \mathbf{x} _i = \mathbf{b} \rightarrow \alpha_j = \frac{\langle \mathbf{x}_j , \space \mathbf{b} \rangle}{|| \mathbf{x}_j ||^2 _2} $$
The span of a set $S = \{ \mathbf{x}_1 \space \dots \space \mathbf{x} _k \}$ is the set of all linear combinations containing its elements.
$$ \text{span} \{S \} = \{\space \mathbf{b} \space | \space \mathbf{b} =\sum _{i = 1} ^k \alpha _i \mathbf{x}_i \hspace{10pt} \alpha_1 \dots \alpha_k \in F \space \} $$
The vector $\mathbf{0}$ exists in any span (simply choose $\alpha_1 = \dots = \alpha _k = 0$ ) and $\text{span} \{ \empty \} = \{ \mathbf{0} \}$.
A vector $\mathbf{b} \in \text{span} \{S\}$ is said to be spanned uniquely by $S$ if $\mathbf{b}$ can be written as exactly one linear combination of elements of $S$.
$$ S \text{ spans all } \mathbf{b} \in \text{span} \{S\} \text{ uniquely} \leftrightarrow S \text{ spans } \mathbf{0} \text{ uniquely} $$
If $S$ spans $\mathbf{0}$ uniquely, then it is also linearly independent, as there would only be one linear combination of its elements which results in $\mathbf{0}$, the one in which all coefficients are $0$.